Question

In the given figure, $ABCD$ is a quadrilateral inscribed in a circle, Diagonals $AC$ and $BD$ are Joined. If $\angle CAD={55}^{\circ }$ and $\angle BDC={20}^{\circ }$ , then$\angle BCD$ will be

Answer 1

$\angle CAD=\angle CBD={55}^o$ (Angles is the same segment)

In $\Delta BCD$,

$\angle BCD+\angle CBD+\angle BDC={180}^o$

$\angle BCD+{55}^o+{20}^o={180}^o$

$\angle BCD+{75}^o={180}^o$

$\angle BCD={180}^o-{75}^o$

$\therefore \angle BCD={105}^o$.